6 L6: Substitution Methods (2.4)
6.1 Definition: Homogeneous Equations
A First-order linear diffyQ
\[\frac{dy}{dx}=f(x,y)\] is said to be homogeneous if \(f(x,y)\) can be expressed as a function \(g(y/x)\)
6.1.1 Example (Level Newbie)
\[\frac{dy}{dx}=\frac{x+y}{x}\]
If we can’t separate1 the diffyQ, then we can make a variable substitution, such that \(f(x,y)\) can be expressed as a function \(g(y/x)\), and this change will make it separable, e.g.
\[\begin{align} \frac{dy}{dx}&=\frac{x+y}{x}\\ &=\frac{x}{x}+\frac{y}{x}\\ &=1+\frac{y}{x}\\ &=g(y/x) \end{align}\]
Because I can write the right side as function of \(y/x\), then a simple substitution, will transform my diffyQ to be separable. \[\begin{align} \text{If }u &=\frac{y}{x}\\ \Rightarrow\\ y&=ux\\ \Rightarrow\\ \frac{dy}{dx}&= u'x+ux'=u'x+u\\ \Rightarrow\\ \text{(Putting this in our diffyQ)}&:\\ \frac{dy}{dx}=u'x+u&=1+u\\ \Rightarrow\\ u'x&=1\\ \frac{du}{dx}x&=1\\ du&=\frac{dx}{x}\qquad (\text{Hurrah! it is separable!})\\ u&=\ln |x|+C\\ \frac{y}{x}&=\ln |x|+C \end{align}\]
\[\boxed{y= x\ln |x| +Cx}\]
6.1.2 General Method
\[\frac{dy}{dx}=g(y/x)\] Let \(u =y/x\), then \(y=ux \rightarrow y'=u'x+u\):
\[\begin{align} \frac{dy}{dx}=u'x+u&=g(u)\\ u'x&=g(u)-u\\ \frac{du}{dx}x&=g(u)-u\\ \frac{du}{g(u)-u}&=\frac{dx}{x}\qquad\text{hurrah!}\\ \int \frac{du}{g(u)-u}&=\int \frac{dx}{x}\\ \Rightarrow \\ \text{Integrate and substitute back.} \end{align}\]
6.1.3 Example (F Level 2)
\[ x\frac{dy}{dx}=y \ln y- y \ln x \] Sol:
To determine if it’s a first-order linear differential equation, let’s examine its characteristics.
First-Order Differential Equation: It is a first-order differential equation since it involves the first derivative of \(y\) with respect to \(x\) and does not involve higher derivatives.
Linearity: A first-order linear differential equation has the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x)\) and \(Q(x)\) are functions of \(x\) only. In your equation, the term \(y\ln y\) is nonlinear because it involves the logarithm of the dependent variable \(y\). Therefore, the given equation is not linear.
\[\begin{align} x(u'x+u)&=ux(\ln (ux)-\ln x)\\ (u'x+u)&=u\ln \left(\frac{(ux)}{x}\right)\\ u'x+u&=u\ln u\\ u'x&=u\ln u -u\\ \frac{du}{dx}x&=u(\ln u -1)\\ \frac{du}{u(\ln u -1)}&=\frac{dx}{x} \end{align}\]
RHS simple trivial integration, LHS, can be solve with a simple substitution:
\[v=(\ln u -1)\rightarrow dv=\frac{1}{u}du\] \[\begin{align} \Rightarrow\\ \frac{du}{u(\ln u -1)}&=\frac{dx}{x}\\ \frac{dv}{v}&=\frac{dx}{x}\\ \ln |v| &= \ln x+C\\ \ln |\ln u -1|&= \ln x +C\\ \ln |\ln y/x -1|&= \ln x +C\\ \ln |\ln y/x -1|&= \ln x +\ln C\\ \ln |\ln y/x -1|&= \ln xC\\ |\ln y/x -1| &= Cx, \qquad \text{x>0}\\ \ln y/x &= 1+Cx\\ y/x &= e^{1+Cx}\\ y&= xe\cdot e^{Cx} \end{align}\]
6.2 Bernoulli Equations
An equation of the form \[ y' +P(x)y=Q(x)y^n\]
is called a Bernoulli equation
6.2.1 General Method
When \(n\neq 0, n\neq 1\) substitute \[\boxed{V=y^{1-n}}\Rightarrow\boxed{\frac{dV}{dx}=(1-n)y^{-n}y'}\]2 \[\begin{align} \Rightarrow\\ y' +P(x)y&=Q(x)y^n\\ \text{mult. by: }& y^{-n}\\ y^{-n}y'+P(x)y^{1-n}&=Q(x)\\ \text{mult. by: }& (1-n)\\ (1-n)y^{-n}y'+(1-n)P(x)y^{1-n}&=(1-n)Q(x)\\ \frac{dV}{dx}+(1-n)P(x)V&=(1-n)Q(x)\\ (\text{Linear!})\\ \Rightarrow\\ \text{Solve }\\ $\Rightarrow$\\ \text{Substitute back.} \end{align}\]
6.2.2 Example (P Level 3)
\[\begin{align} \cos x\frac{dy}{dx}-y\sin x+y^2&=0\\ \frac{dy}{dx}-(\tan x)y +(\sec x)y^2&=0\\ y' +(-\tan x)\,y&=(-\sec x)y^2 \end{align}\]
\[\boxed{\boxed{n=2}\rightarrow V=y^{1-2}=y^{-1}\rightarrow \frac{dV}{dx}=-y^{-2}y'}\] \[\begin{align} y' +(-\tan x)\,y&=(-\sec x)y^2\\ \text{(Mult. by: )}& -y^{-2} \\ -y^{-2}y' +(-y^{-2})(-\tan x)y&=(-y^{-2})(-\sec x)y^2\\ \frac{dV}{dx}+y^{-1}\tan x&=\sec x\\ \frac{dV}{dx}+(\tan x) V&=\sec x\\ \text{(Linear)}\\ \cdots \end{align}\]
6.3 Definition: The nonlinear equation
The nonlinear equation \[\boxed{\frac{dy}{dx}+P(x)y=Q(x)y^2+R(x)}\] is known as the Riccoti equation.
Note: If \(R(x)=0\) \(\Rightarrow\) quadratic Bernoulli
6.3.1 General Method
Suppose we know a particular solution \[y=y_1(x)\]
Then substitute \[\boxed{y=y_1+\frac{1}{u}}\] to find a general sol\(^{n}\)
\[\begin{align} y'&=y_1'-\frac{1}{u^2}u'\\ \Rightarrow\\ y_1'-\frac{u'}{u^2} +P(x)\left(y_1+\frac{1}{u}\right)&=Q(x)\left(y_1+\frac{1}{u}\right)^2+R(x)\\ \text{We know that }&y_1\text{ is a soln:} \\ y_1'+Py_1&=Qy_1^2+R\\ \Rightarrow\\ -\frac{u'}{u^2}+P(x)\frac{1}{u}&=Q(x)\left[\frac{2}{u}y_1+\frac{1}{u^2}\right]\\ u'-P(x)u&=-Q(x)[2uy_1+1]\\ u'+[2y_1Q(x)-P(x)]u&=-Q(x)\\ \text{(Linear!!)}\\ \Rightarrow \end{align}\]
Solve and substitute back!
6.3.2 Example
\[\boxed{y'+2y+y^2=0,\qquad y_1=-2}\]
We assume that a particular + \(\frac{1}{u}\) is also a solution:
\[\boxed{y=-2+\frac{1}{u}\rightarrow \frac{dy}{dx}=-\frac{u'}{u^2}}\]
\[\begin{align} -\frac{u'}{u^2}+2\left(-2+\frac{1}{u}\right)+\left(-2+\frac{1}{u}\right)^2&=0\\ -\frac{u'}{u^2}-4+\frac{2}{u}+4-\frac{4}{u}+\frac{1}{u^2}&=0\\ -\frac{u'}{u^2}+\frac{2}{u}-\frac{4}{u}+\frac{1}{u^2}&=0\\ -u'+2u-4u+1&=0\\ u'+2u&=-1\\ \text{Linear}\\ \cdots \end{align}\]
6.4 One more method for substitution, partial derivatives.
Example:
\[\frac{dy}{dx}=(x+y+3)^2\]
a convenient substitution for that equations will be:
\[v=x+y+3\] \[\rightarrow\] \[y=\beta(x,v)\] \[\rightarrow\] \[\frac{dy}{dx}=\beta_x\frac{dx}{dx}+\beta_v\frac{dv}{dx}\]
6.4.1 Solution
\[\beta = v-x-3\] \[\Rightarrow\] \[\beta_x = -1\] \[\beta_v = 1\] \[\Rightarrow\] \[\frac{dy}{dx}=(x+y+3)^2, \text{BECOMES!}\]
\[-1+1\frac{dv}{dx}=(v)^2\]
\[\frac{dv}{dx}=1+v^2\]
\[\frac{dv}{1+v^2}=dx\]
\[\cdots\]
6.5 Summary
- Homogeneous: \[\boxed{\frac{dy}{dx}=f(x,y)}\rightarrow \boxed{\frac{dy}{dx}=g(y/x)}\]
- Bernoulli: \[\boxed{y' +P(x)y=Q(x)y^n}\rightarrow \boxed{V=y^{1-n}}\]
- Nonlinear, Riccoti: \[\boxed{\frac{dy}{dx}+P(x)y=Q(x)y^2+R(x)}\rightarrow \boxed{y=y_1+\frac{1}{u}}\]
6.6 References
- Khan: Homogeneous diffyQ https://www.khanacademy.org/math/differential-equations/first-order-differential-equations#homogeneous-equations
- Wiki: Homogeneous diffyQ https://en.wikipedia.org/wiki/Homogeneous_differential_equation
- Math is fun: examples of Homogeneous diffyQ https://www.mathsisfun.com/calculus/differential-equations-homogeneous.html
- Khan, exact diffyQ (intuition 1) https://youtu.be/iEpqcdaJNTQ
- Khan, exact diffyQ (intuition 2) https://youtu.be/a7wYAtMjORQ
- Quarto revealjs. https://quarto.org/docs/presentations/revealjs/themes.html